Coxeter Complexes
3.6 Products and Convexity
3.6.4 Semigroup Structure
GivenA, B∈Σ, we wish to define theirproductABas in Section 1.4.6. Thus AB should be the simplex with sign sequence given by
σH(AB) =
σH(A) ifσH(A)= 0,
σH(B) ifσH(A) = 0, (3.9) for any wall H ∈ H. Of course, one has to prove the existence of such a simplex.
The existence proof was quite easy in the setting of Chapter 1, and we have already given the analogous easy proof in general using the Tits cone (Proposition 2.82, part (1)). But we will give an independent proof here, which is purely combinatorial. It is longer, but it is instructive, and it will generalize to buildings in the next chapter. In the course of the proof we will see that the chambers having AB as a face are precisely those that can start a minimal gallery fromAtoB, as we would expect from Exercise 1.62 [and Section 2.7].
This completely characterizes the desiredAB, since a simplex is determined by the set of chambers having it as a face; see, for instance, Corollary 3.17 or Exercise 3.10.
The existence of the desiredAB is quite easy to prove ifB is a chamber (in which case (3.9) forcesAB to be a chamber), so we begin with that case.
The result should be compared with Proposition 1.40.
3.6 Products and Convexity 151 Proposition 3.103.Given a simplexAand a chamberC, there is a (unique) chamberAC such that for anyH∈ H,
σH(AC) =
σH(A) ifσH(A)= 0,
σH(C) ifσH(A) = 0. (3.10) It has A as a face, and among the chambers having A as a face, it is the unique one at minimal distance from C. Every minimal gallery from Ato C starts withAC.
Proof. Choose a minimal gallery Γ: C0, . . . , Cd = C from A to C. Given H ∈ H, ifσH(A) = 0, thenσH(C0) =σH(A), sinceC0 ≥A. If σH(A) = 0, thenH /∈ S(A, C) [see Definition 3.77], so Proposition 3.78 implies thatΓ does not crossH; henceσH(C0) =σH(C). ThusC0is the desired chamberAC. The last two assertions of the proposition follow from the existence proof, since we started with an arbitrary minimal gallery fromAtoC. Alternatively, the second-to-last assertion follows from (3.10) as in the proof of Proposition 1.40, and the last assertion is simply a restatement of it.
Definition 3.104.Given simplicesA, C∈ΣwithCa chamber, theirproduct is the chamberACdescribed in Proposition 3.103. The product is also denoted by projAC and called theprojection ofC ontoA.
Equation (3.10) leads to the following important property of AC, which we call thegate property.
Proposition 3.105.For any simplexAand any chambersC, DwithD≥A, d(C, D) =d(C, AC) +d(AC, D). (3.11) Proof. Partition the wallsH separatingCfromDinto two subsets according to whether or not σH(A) = 0. Those withσH(A) = 0 are precisely the walls separating AC fromD, while those withσH(A)= 0 are the walls separating AC from C. Equation (3.11) now follows from the fact that we can compute distances by counting separating walls (Lemma 3.69).
We already saw the gate property in the context of hyperplane arrange- ments in Chapter 1; see Exercise 1.42 and Figure 1.6. As we noted there, it says that C≥A is a “gated subset” of the metric space C=C(Σ) in the sense of Dress–Scharlau [97].
We now proceed to the existence ofABfor arbitraryB. We will use ideas borrowed from the Dress–Scharlau theory, with some simplifications achieved in the present context by the use of sign sequences. Our first goal will be to find a simplexABsuch that the chambers≥ABare precisely those that can start a minimal gallery fromAtoB. We will then be able to check that (3.9) holds.
Let C = C(Σ), and for any simplex A, let CA ⊆ C be the set C≥A of chambers ≥ A. For any two simplices A, B, let CA,B ⊆ CA be the set of chambers that can start a minimal gallery from AtoB.
Proposition 3.106.
(1)CA,B is the image of the projection mapCB→ CA given by D→AD.
(2)The projection maps CA,B CB,A, given by C → BC and D → AD (C∈ CA,B, D∈ CB,A), define mutually inverse bijections.
(3)Given any minimal gallery C0, . . . , Cl from A to B, we have C0 =ACl and Cl = BC0. In other words, C0 and Cl correspond to one another under the bijections in (2).
Proof. A minimal galleryC0, . . . , ClfromAtoBis also minimal fromAtoCl, so C0=ACl by Proposition 3.103. Similarly,Cl=BC0. This proves (3) and shows that CA,B is contained in the image of the projection map CB → CA, which is part of (1). To prove the opposite inclusion, consider anyD ∈ CB. Then one sees by checking sign sequences that S(AD, B(AD)) = S(A, B);
henced(AD, B(AD)) =d(A, B) by Proposition 3.78. Thus there is a minimal gallery from A to B starting with AD. This proves (1). It follows from (1) that the projection maps do define mapsCA,BCB,A, and it is easy to check (using sign sequences) that these maps are inverse to one another, whence (2).
We need one more simple observation before we can complete the analysis ofCA,B.
Lemma 3.107.The projectionC → CA takes adjacent chambers to chambers that are equal or adjacent. Consequently, CA,B is a connected subset of the chamber graph of Σ, i.e., any two elements of CA,B can be connected by a gallery inCA,B.
Proof. The first assertion is immediate if one calculates projections in terms of sign sequences (equation (3.10)). The second assertion now follows from part (1) of Proposition 3.106 becauseCB is connected (by Proposition 3.16 or
Proposition 3.93).
We can now prove the main result of this subsection. Choose a type function on Σ with values in a set S, so that we can define the Weyl group W = WM and the Weyl distance functionδ as in Section 3.5. Recall that the abstract setS then becomes a set of generators ofW.
It will be convenient to use residue terminology in what follows. Recall that by Corollary 3.17, the residues are the sets of the form CA, one for each simplex A. The main point in what follows is to show that CA,B is again a residue, so that we can defineAB to be the corresponding simplex.
Theorem 3.108.Let A be a simplex of cotype J, let B be a simplex of co- typeK, and letw=δ(A, B). ThenCA,Bis a residue of typeJ1:=J∩wKw−1. In other words, there is a simplex AB of cotype J1 such that CA,B = CAB, i.e., the chambers that can start a minimal gallery fromA toB are precisely those having AB as a face. Moreover, the sign sequence of AB is given by equation (3.9).
3.6 Products and Convexity 153 Proof. To show thatCA,Bis contained in a residue of typeJ1it suffices, by the lemma, to prove that any two adjacent chambers in CA,B areJ1-equivalent.
Let C1, C2 ∈ CA,B be s-adjacent, and let D1 = BC1 and D2 = BC2. Then D1 andD2 are distinct by part (2) of Proposition 3.106 and are adjacent by Lemma 3.107. Let t∈S be the element such thatD1 andD2 aret-adjacent.
For i = 1,2 there is a minimal gallery from A to B starting with Ci and, necessarily, ending withDi(see part (3) of Proposition 3.106). Soδ(C1, D1) = w=δ(C2, D2). Computingδ(C1, D2) in two different ways, we conclude that sw = wt. See Figure 3.7, which should be viewed as a schematic picture of the chamber graph. We haves∈J because C1, C2≥A, and similarly t∈K.
C1
D2 w
w t
D1
CB
CA s
C2
Fig. 3.7.sw=wt.
Sos=wtw−1∈J∩wKw−1, whence C1 andC2areJ1-equivalent.
To show thatCA,B is an entire residue of typeJ1, it suffices to prove that ifC1is inCA,BandC2iss-adjacent toC1for somes∈J1, thenC2is inCA,B. Writesw =wt with t∈ J, and let D1 =BC1. Thenδ(C2, D1) =sw =wt;
see Figure 3.8. Since decompositions ofδ(C2, D1) correspond to galleries from
C1 w
D1
CB
CA
C2
s
Fig. 3.8.δ(C2, D1) =sw.
C2 toD1, it follows that there is a chamberD2 such thatδ(C2, D2) =wand δ(D2, D1) =t. In other words, we have achieved the situation in Figure 3.7, where D2 ∈ CB becauset ∈K. Sinced(C2, D2) = l(w) = d(A, B), it follows that C2∈ CA,B, as required.
Now letABbe the simplex such thatCA,B=CAB. We must calculate the sign sequence ofAB. We haveAB≥AsinceCAB⊆ CA; soσH(AB) =σH(A) if σH(A) = 0. Suppose σH(A) = 0. Then H /∈ S(A, B), so the minimal galleriesΓ:C0, . . . , Cl fromAtoBdo not crossH, i.e.,C0 andClare on the
same side ofH. IfσH(B)= 0, it follows thatσH(C0) =σH(B). In other words, every chamber C ≥AB satisfiesσH(C) =σH(B); hence σH(AB) =σH(B) by Exercise 3.81. If σH(B) = 0, on the other hand, then Γ is in one root α associated to H, and we can fold onto −α to get another minimal gallery fromAtoB. Thus there are elements ofCA,B=CAB on both sides ofH, and
σH(AB) = 0. This proves (3.9).
Definition 3.109.Given simplices A, B ∈ Σ, their product is the cham- berABdescribed in Theorem 3.108 and characterized by equation (3.9). The product is also denoted by projAB and called theprojection ofB ontoA.
As in Section 1.4.6, equation (3.9) has the following consequence:
Corollary 3.110.The product of simplices is associative. HenceΣis a semi-
group.
Example 3.111.LetCandCbe adjacent chambers. Letv, vbe the vertices of C, C that are not in the common panel, as in Figure 3.9. Consider the
v
C C
v
Fig. 3.9.Adjacent chambers.
product vv. We show by two different methods that vv ≤ C. Method 1:
There is a minimal galleryC, C from v tov sincevand v are not joinable.
[They have the same type.] Hence vv is a face of the starting chamber C.
Method 2: Use sign sequences. Assume for simplicity (and without loss of generality) that σH(C) = + for every wall H, so that σH(v)≥0 for all H.
We then haveσH(C) = + for allH except the one containing P :=C∩C; henceσH(v)≥0 for allH except the one containingP. SinceσH(v) = + for that exceptional wall, it follows that σH(vv) ≥ 0 for all H and hence that vv≤C.
We close this subsection by recording some connections between the poset and semigroup structures onΣas in Proposition 1.41 and Exercise 1.44. The proofs are easy via sign sequences and are left to the reader
Proposition 3.112.LetA andB be arbitrary simplices in Σ.
(1)A≤AB, with equality if and only ifsuppB ≤suppA.
3.6 Products and Convexity 155 (2)A≤B if and only ifAB=B.
(3) suppA= suppB if and only if AB=A andBA=B.
(4)IfsuppA= suppB, then left multiplication byB andAdefines mutually inverse bijections Σ≥AΣ≥B. In particular,dimA= dimB.
(5)ABandBAhave the same support, which is the intersection of the walls
containing both AandB.
Corollary 3.113.For any simplices A, B ∈ Σ, dimAB = dimBA. Conse- quently,dimAB≥max{dimA,dimB}.
Proof. The first assertion follows immediately from parts (5) and (4) of the proposition. For the second, we have dimAB≥dimAtrivially because A≤ AB, and similarly dimBA≥dimB; now use the fact that dimBA= dimAB.
Exercises
3.114.Use Theorem 3.108 to give a new proof of Lemma 2.25.
3.115.Show that every root is a subsemigroup, and hence every intersection of roots is a subsemigroup. In particular, this applies to the support of any simplex (Definition 3.98).
3.116.Show that (finitely many) simplices A, B, . . . , C are joinable if and only if they commute with one another in the semigroup Σ, in which case their product is their least upper bound (see Exercise 1.43). Deduce, as in the proof of Proposition 1.127, thatΣ is a flag complex.
3.117.Given simplices A1, A2, B ∈Σ with A1 ≤A2, show thatd(A1, B)≤ d(A2, B), with equality if A2 ≥A1B. In particular, d(A, B) =d(AB, B) for any two simplicesA, B.
3.118.Figure 3.9 suggests thatvv=C. Give examples to show that this is not necessarily the case. For instance,vv could be a vertex or an edge.
3.119.Recall that the linkLA:= lkΣA of any simplexAis again a Coxeter complex; hence it has a semigroup structure. Is it a subsemigroup of Σ? If not, how is the product on LA related to the product inΣ?
The remaining exercises are intended to show how the use of products can sometimes replace arguments based on the Tits cone. The intent of the exercises, then, is that they should be solved combinatorially, without the Tits cone. Given a chamber C and a panelP of C, the wall containingP will be called awall of C. Thus every chamber has exactly n+ 1 walls if dimΣ=n.
3.120.Fix a chamberC and letHC be its set of walls.
(a) Show that C is defined by HC; in other words, if D is a chamber such that σH(D) =σH(C) for all H∈ HC, thenD=C.
(b) SupposeAis a simplex such thatσH(A)≤σH(C) for allH∈ HC. Show that A≤C.
(c) IfAandBare faces ofC, show thatA≤Bif and only ifσH(A)≤σH(B) for allH ∈ HC.
(d) IfA≤C, show thatAis defined byHC; in other words, ifB is a simplex such thatσH(B) =σH(A) for allH ∈ HC, thenB=A.
3.121. (a) LetCbe a chamber, and letsandtbe reflections with respect to two distinct walls ofC, denoted byHsandHt. Letmbe the order ofst, and assumem≥3. IfD is another chamber that also hasHsandHtas two of its walls, show that eitherHsandHtboth separateC fromD or else neither of them separatesC fromD.
(b) Give an example to show that we cannot drop the assumption thatm≥3 in (a).
(c) Generalize (a) as follows. Let H1, . . . , Hk be walls of a chamber C such that the corresponding reflections si generate an irreducible Coxeter group. If D is another chamber having H1, . . . , Hk as walls, show that either everyHi separatesC fromD or else no Hi separatesC fromD.
3.122.Use the previous exercise to give a combinatorial proof of the following fact, which we have proven earlier by different methods: If (W, S) is irreducible andwSw−1=S for some w= 1 in W, thenW is finite andw is the longest element. [We gave two proofs of this for finite reflection groups, one algebraic and one geometric; see the proof of Corollary 1.91. And we generalized the algebraic proof to the infinite case in the proof of Proposition 2.73. The point of the exercise is that we now have the tools to generalize the geometric proof.]